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To determine the number of ways to select 3 books from a group of 11 books, we need to use the combination formula. The combination formula is given by [tex]\({ }_n C _r = \frac{n!}{r!(n-r)!}\)[/tex], where [tex]\(n\)[/tex] is the total number of items to choose from, and [tex]\(r\)[/tex] is the number of items to be chosen.
Here, [tex]\(n = 11\)[/tex] and [tex]\(r = 3\)[/tex].
1. First, we identify [tex]\(n\)[/tex] and [tex]\(r\)[/tex]:
[tex]\[ n = 11, \quad r = 3 \][/tex]
2. Next, we substitute [tex]\(n\)[/tex] and [tex]\(r\)[/tex] into the combination formula:
[tex]\[ { }_{11}C_{3} = \frac{11!}{3!(11-3)!} \][/tex]
3. We simplify the denominator by calculating [tex]\(11 - 3\)[/tex]:
[tex]\[ 11 - 3 = 8 \][/tex]
4. The formula now becomes:
[tex]\[ { }_{11}C_{3} = \frac{11!}{3! \cdot 8!} \][/tex]
5. We observe that [tex]\(11!\)[/tex] can be expanded and simplified to cancel out the [tex]\(8!\)[/tex] in the numerator and denominator:
[tex]\[ 11! = 11 \times 10 \times 9 \times 8! \][/tex]
[tex]\[ \frac{11 \times 10 \times 9 \times 8!}{3! \times 8!} \][/tex]
6. The [tex]\(8!\)[/tex] in the numerator and the denominator cancel each other out:
[tex]\[ \frac{11 \times 10 \times 9}{3!} \][/tex]
7. We calculate [tex]\(3!\)[/tex]:
[tex]\[ 3! = 3 \times 2 \times 1 = 6 \][/tex]
8. Now, we have:
[tex]\[ \frac{11 \times 10 \times 9}{6} \][/tex]
9. Performing the multiplication on the numerator:
[tex]\[ 11 \times 10 = 110 \][/tex]
[tex]\[ 110 \times 9 = 990 \][/tex]
10. We then divide by the denominator:
[tex]\[ \frac{990}{6} = 165 \][/tex]
Therefore, the number of ways to select 3 books from a group of 11 books is:
[tex]\[ 165 \][/tex]
Here, [tex]\(n = 11\)[/tex] and [tex]\(r = 3\)[/tex].
1. First, we identify [tex]\(n\)[/tex] and [tex]\(r\)[/tex]:
[tex]\[ n = 11, \quad r = 3 \][/tex]
2. Next, we substitute [tex]\(n\)[/tex] and [tex]\(r\)[/tex] into the combination formula:
[tex]\[ { }_{11}C_{3} = \frac{11!}{3!(11-3)!} \][/tex]
3. We simplify the denominator by calculating [tex]\(11 - 3\)[/tex]:
[tex]\[ 11 - 3 = 8 \][/tex]
4. The formula now becomes:
[tex]\[ { }_{11}C_{3} = \frac{11!}{3! \cdot 8!} \][/tex]
5. We observe that [tex]\(11!\)[/tex] can be expanded and simplified to cancel out the [tex]\(8!\)[/tex] in the numerator and denominator:
[tex]\[ 11! = 11 \times 10 \times 9 \times 8! \][/tex]
[tex]\[ \frac{11 \times 10 \times 9 \times 8!}{3! \times 8!} \][/tex]
6. The [tex]\(8!\)[/tex] in the numerator and the denominator cancel each other out:
[tex]\[ \frac{11 \times 10 \times 9}{3!} \][/tex]
7. We calculate [tex]\(3!\)[/tex]:
[tex]\[ 3! = 3 \times 2 \times 1 = 6 \][/tex]
8. Now, we have:
[tex]\[ \frac{11 \times 10 \times 9}{6} \][/tex]
9. Performing the multiplication on the numerator:
[tex]\[ 11 \times 10 = 110 \][/tex]
[tex]\[ 110 \times 9 = 990 \][/tex]
10. We then divide by the denominator:
[tex]\[ \frac{990}{6} = 165 \][/tex]
Therefore, the number of ways to select 3 books from a group of 11 books is:
[tex]\[ 165 \][/tex]
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